Question

(4x+y)^2dx/dy=1 Solve the differential equation​

Answer 1

Answer:

$\frac{dy}{dx} = (4x + y) {}^{2}$

Let 4x + y = t

4 + dy/dx = dt/dx

dy/dx = dt/dx - 4

$\frac{dt}{dx} - 4 = {t}^{2}$

$\frac{dt}{dx} = {t}^{2} + 4$

$\frac{dt}{ {t}^{2} + 4 } = dx$

Integrating both sides, we get

$\tan {}^{ - 1} ( \frac{t}{2} ) = x + c$

Replacing t, we get..

$\tan {}^{ - 1} ( \frac{4x + y}{2} ) = x + c$

Ans.

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